Quiz for fun, I/O pin and 4 levels

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KerimF
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Posted by KerimF: Thu. Jul 11, 2019 - 05:59 PM
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Let us assume; we like to monitor on a scope the timing of 4 stages while an MCU code is running. But one I/O pin only is free for this task.

 

As we know, an I/O pin (as of ATmega8) is controlled by 2 internal bits (in two I/O registers; DDRx and PORTx). An I/O pin could have, therefore, 4 different states which could be made to output four different voltages. The two obvious outputs are Vcc and 0V.

 

On my next post, I will present a possible solution which I applied lately.

 

Have fun,

Cheers,

 

Kerim

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Learning and Information, General Electronics
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ki0bk
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Posted by ki0bk: Thu. Jul 11, 2019 - 06:07 PM
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Well one solution, and I did not think of this but read it here somewhere, is to have a soft uart (TX only) and send serial debug info out as needed and read the result using a logic analyzer......

I'm sure there are others as well.

 

Jim

 

 

Keys to wealth:

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Brian Fairchild
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Posted by Brian Fairchild: Thu. Jul 11, 2019 - 06:36 PM
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Two external resistors, one pull-up and one pull-down.

 

With the pin as an output you have +Vcc and 0V. As as input you can turn on and off the internal pull-up and therefore have two intermediate levels.

 

With a bit of maths you could arrange the intermediate levels to be 1/3Vcc and 2/3Vcc.

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#2 Hardware Problem? Read AVR042.

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

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KerimF
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Posted by KerimF: Thu. Jul 11, 2019 - 06:36 PM
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Hi Jim,

The solution, you presented, is indeed good when there are too many stages to monitor, not just four.

 

Hint:

For 4 states, it could be solved by a simple hardware; so that the pin voltage changes as fast as setting/clearing its DDR or PORT controlling bit.

 

    

Last Edited: Thu. Jul 11, 2019 - 06:40 PM
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KerimF
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Posted by KerimF: Thu. Jul 11, 2019 - 06:38 PM
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Hi Brian,

 

You were very fast crying

 

 

 A possible solution:

Two external resistors are added; a pull-up (R_up) and a pull-down (R_dw).

In this case, the optimum levels for monitoring are: Vcc , 2*Vcc/3 , Vcc/3 and 0

 

If Rp is the typical value of the internal pull-up resistor, the 4 levels of the pin output are:

DDR=1 and PORT=1 ===> Vcc

DDR=0 and PORT=1 ===> Vcc * R_dw / ( R_dw + Rp//R_up )

DDR=0 and PORT=0 ===> Vcc * R_dw / ( R_dw + R_up )

DDR=0 and PORT=0 ===> Vcc * R_dw / ( R_dw + R_up )

DDR=1 and PORT=0 ===> 0

 

By solving the two equations below:

2*Vcc/3 =  Vcc * R_dw / ( R_dw + Rp//R_up )

Vcc/3 = Vcc * R_dw / ( R_dw + R_up )

We get:

R_up = 3*Rp

R_dw = R_up / 2

 

On my board, the typical value of Rp is 37.5 K (of ATmega8; 80uA/3V). I added R_up (120K) and R_dw( 56K).

DDR=1 and PORT=1 ===> 5V00

DDR=0 and PORT=1 ===> 3V31

DDR=0 and PORT=0 ===> 1V59

DDR=0 and PORT=0 ===> 1V59

DDR=1 and PORT=0 ===> 0V00

 

 

Edited: Thanks to Jim.

 

Last Edited: Fri. Jul 12, 2019 - 07:28 AM
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Brian Fairchild
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Posted by Brian Fairchild: Thu. Jul 11, 2019 - 06:39 PM
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If we assume the internal pull-up is 100k then an external pull-up of 200k, and a pull-down of 100k, will give you then numbers above.

 

Assuming 5v rail.

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#2 Hardware Problem? Read AVR042.

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

Last Edited: Thu. Jul 11, 2019 - 06:40 PM
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Brian Fairchild
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Posted by Brian Fairchild: Thu. Jul 11, 2019 - 06:41 PM
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There are other possibility depending on what pin you have available. One option might be to change the reload value of a timer if a timer output pin is free and it's not otherwise used internally.

 

[E2A]

Going back many years an ex-colleague put an R-2R-4R... resistor DAC on the address bus of his Z-80 based CP/M system and connected it to a moving coil voltmeter. The meter therefore indicated where in the address range the code was. A low voltage would indicate user code whereas a high voltage either somewhere in the BDOS or BIOS.

#1 Hardware Problem? https://www.avrfreaks.net/forum/...

#2 Hardware Problem? Read AVR042.

#3 All grounds are not created equal

#4 Have you proved your chip is running at xxMHz?

#5 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand."

Last Edited: Thu. Jul 11, 2019 - 06:44 PM
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KerimF
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Posted by KerimF: Thu. Jul 11, 2019 - 06:45 PM
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If we assume the internal pull-up is 100k then an external pull-up of 200k, and a pull-down of 100k, will give you then numbers above.

 

I am afraid that in your case (internal pull-up is 100K), the external pull-up is 300K and the pull-down is 150K.

 

 

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ki0bk
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Posted by ki0bk: Thu. Jul 11, 2019 - 07:00 PM
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KerimF wrote:

DDR=0 and PORT=0 ===> Vcc * R_dw / ( R_dw + R_up )

DDR=0 and PORT=0 ===> 0

This does not look right?  both ddr/port = 0 in both cases???

 

 

Jim

edit: I think you meant:

DDR=1 and PORT=0 ===> 0

 

Keys to wealth:

Invest for cash flow, not capital gains!

Wealth is attracted, not chased! 

Income is proportional to how many you serve!

If you want something you've never had...

...you must be willing to do something you've never done!

Lets go Brandon!

Last Edited: Thu. Jul 11, 2019 - 07:05 PM
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KerimF
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Posted by KerimF: Thu. Jul 11, 2019 - 07:09 PM
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Thank you Jim.

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ki0bk
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Posted by ki0bk: Thu. Jul 11, 2019 - 07:32 PM
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KerimF wrote:

DDR=1 and PORT=1 ===> 5V00

DDR=0 and PORT=1 ===> 3V31

DDR=1 and PORT=0 ===> 1V59

DDR=0 and PORT=0 ===> 1V59

DDR=0 and PORT=0 ===> 0V00

 

This still does not look right?  One more time please.

 

 

Keys to wealth:

Invest for cash flow, not capital gains!

Wealth is attracted, not chased! 

Income is proportional to how many you serve!

If you want something you've never had...

...you must be willing to do something you've never done!

Lets go Brandon!

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John_A_Brown
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Posted by John_A_Brown: Fri. Jul 12, 2019 - 06:53 AM (Reply to #7)
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Also many years ago, I have seen a similar thing done with a scope, using upper 8 bit of address bus as one axis, and lower 8 for the other.
I can only assume it wasn't all that useful, as I've not seen it suggested since.

Four legs good, two legs bad, three legs stable.

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N.Winterbottom
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Posted by N.Winterbottom: Fri. Jul 12, 2019 - 07:18 AM
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ki0bk wrote:
Well one solution, and I did not think of this but read it here somewhere, is to have a soft uart (TX only) and send serial debug info out as needed and read the result using a logic analyzer......

Originally by Jack Ganssle I bellieve

http://www.ganssle.com/tem/tem226.htm

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John_A_Brown
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Posted by John_A_Brown: Fri. Jul 12, 2019 - 09:39 AM
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N.Winterbottom wrote:

ki0bk wrote:
Well one solution, and I did not think of this but read it here somewhere, is to have a soft uart (TX only) and send serial debug info out as needed and read the result using a logic analyzer......

Originally by Jack Ganssle I bellieve

http://www.ganssle.com/tem/tem226.htm

Really? I'd have thought bit banging serial TX only was as old as the asynch serial protocol.

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sparrow2
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Posted by sparrow2: Fri. Jul 12, 2019 - 11:32 AM
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About where the code run , back in the old days (I think it was late 80's), one of my friends had a PC (80286 I think), that had a LED bar on the case, that showed where the code ran.

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